integral (x(1+3x^2)^5)/2

Lösung

integral

\frac{x ( 1 + 3 x ^{2} ) ^{5}}{2}

\frac{1}{8} (2 x^{2} + 15 x^{4} + 60 x^{6} + 135 x^{8} + 162 x^{10} + 81 x^{12}) + C

Schritte zur Lösung

\int \frac{x ( 1 + 3 x ^{2} ) ^{5}}{2} d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int x (1 + 3 x^{2})^{5} d x

Multipliziere aus

x (1 + 3 x^{2})^{5} : x + 15 x^{3} + 90 x^{5} + 270 x^{7} + 405 x^{9} + 243 x^{11}

= \frac{1}{2} \cdot \int x + 15 x^{3} + 90 x^{5} + 270 x^{7} + 405 x^{9} + 243 x^{11} d x

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int x d x + \int 15 x^{3} d x + \int 90 x^{5} d x + \int 270 x^{7} d x + \int 405 x^{9} d x + \int 243 x^{11} d x)

\int x d x = \frac{x ^{2}}{2}

\int 15 x^{3} d x = \frac{15 x ^{4}}{4}

\int 90 x^{5} d x = 15 x^{6}

\int 270 x^{7} d x = \frac{135 x ^{8}}{4}

\int 405 x^{9} d x = \frac{81 x ^{10}}{2}

\int 243 x^{11} d x = \frac{81 x ^{12}}{4}

= \frac{1}{2} (\frac{x ^{2}}{2} + \frac{15 x ^{4}}{4} + 15 x^{6} + \frac{135 x ^{8}}{4} + \frac{81 x ^{10}}{2} + \frac{81 x ^{12}}{4})

Vereinfache

\frac{1}{2} (\frac{x ^{2}}{2} + \frac{15 x ^{4}}{4} + 15 x^{6} + \frac{135 x ^{8}}{4} + \frac{81 x ^{10}}{2} + \frac{81 x ^{12}}{4}) : \frac{1}{8} (2 x^{2} + 15 x^{4} + 60 x^{6} + 135 x^{8} + 162 x^{10} + 81 x^{12})

= \frac{1}{8} (2 x^{2} + 15 x^{4} + 60 x^{6} + 135 x^{8} + 162 x^{10} + 81 x^{12})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{8} (2 x^{2} + 15 x^{4} + 60 x^{6} + 135 x^{8} + 162 x^{10} + 81 x^{12}) + C

in t e g r a l \frac{2}{t ^{3}}

in t e g r a l \frac{x}{sin ^{2} ( x ^{2} )}

in t e g r a l \frac{x}{16}

in t e g r a l \frac{cos ^{2} ( 2 x )}{sin ( 2 x )}

in t e g r a l - 3 sin (x)